Factor polynomial $$ 7x^{3}-21x^{2}-70x $$

$$ 7x^{3}-21x^{2}-70x = 7x( x+2 )( x-5 ) $$

Step 1 :

After factoring out $ 7x $ we have:

$$ 7x^{3}-21x^{2}-70x = 7x ( x^{2}-3x-10 ) $$

Step 2 :

Step 2: Identify constants $ \color{blue}{ b }$ and $\color{red}{ c }$. ( $ \color{blue}{ b }$ is a number in front of the $ x $ term and $ \color{red}{ c } $ is a constant). In our case:

$$ \color{blue}{ b = -3 } ~ \text{ and } ~ \color{red}{ c = -10 }$$

Now we must discover two numbers that sum up to $ \color{blue}{ -3 } $ and multiply to $ \color{red}{ -10 } $.

Step 3: Find out pairs of numbers with a product of $\color{red}{ c = -10 }$.

PRODUCT = -10
-1     10	1     -10
-2     5	2     -5

Step 4: Find out which pair sums up to $\color{blue}{ b = -3 }$

PRODUCT = -10 and SUM = -3
-1     10	1     -10
-2     5	2     -5

Step 5: Put 2 and -5 into placeholders to get factored form.

$$ \begin{aligned} x^{2}-3x-10 & = (x + \color{orangered}{\square} )(x + \color{orangered}{\square}) \\ x^{2}-3x-10 & = (x + 2)(x -5) \end{aligned} $$

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